At a certain place, the horizontal component of earth's magnetic field is $\frac{1}{\sqrt{3}}$ times the…
At a certain place, the horizontal component of earth's magnetic field is $\frac{1}{\sqrt{3}}$ times the vertical component. The angle of dip at that place is
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
Solution
Given, $B_H / B_V=1 / \sqrt{3}$
where, $B_H$ and $B_V$ are respectively, the horizontal and vertical components of earth's magnetic field
$\begin{aligned}
\tan \delta & =B_V / B_H=\frac{\sqrt{3} B_H}{B_H}=\sqrt{3} \\
\delta & =\tan ^{-1} \sqrt{3}=60^{\circ}
\end{aligned}$
Angle of dip, $\delta=60^{\circ}$